Set Theory Problem: equality of functions

In summary: The converse is clear.So you have shown thatF=G~\Leftrightarrow~dom(F)=dom(G)~\text{and}~F(x)=G(x)~\forall x\in dom(F) I think that was the point of the exercise, maybe it wasn't clear, I don't know. In summary, the statement to be proven is that two functions, F and G, are equal if and only if their domains are equal and their outputs are equal for all inputs in their shared domain. This is shown by first proving that if F is equal to G, then xFy if and only if xGy, and then using this to show that if x is in the domain of
  • #1
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Homework Statement



Prove:

Let F and G be functions. F = G if and only if domF = domG and F(x) = G(x) for all x [tex]\in[/tex] domF.

Homework Equations



?

The Attempt at a Solution



If F = G, then (xFy if and only if xGy) (Substitutivity of identicals)

If (xFy if and only if xGy), then F = G (Axiom of extensionality)

So F = G if and only if (xFy if and only if xGy)

Since F and G are functions,

For all x [tex]\in[/tex] domF, xFy if and only if y = F(x)
For all x [tex]\in[/tex] domG, xGy if and only if y = G(x)

xFy if and only if x [tex]\in[/tex] domF and xFy
xGy if and only if x [tex]\in[/tex] domG and xGy

xFy if and only if x [tex]\in[/tex] domF and y = F(x)
xGy if and only if x [tex]\in[/tex] domG and y = G(x)

So

F = G if and only if ((x [tex]\in[/tex] domF and y = F(x)) if and only if (x [tex]\in[/tex] domG and y = G(x)))

which is equivalent to

F = G if and only if ((x [tex]\in[/tex] domF and x [tex]\in[/tex] domG and y = F(x) and y = G(x)) or (x [tex]\in[/tex] domF and x [tex]\in[/tex] domG and y [tex]\neq[/tex] F(x) and y [tex]\neq[/tex] G(x)) or (x [tex]\notin[/tex] domF and x [tex]\notin[/tex] domG) or (x [tex]\in[/tex] domF and x [tex]\notin[/tex] domG and y [tex]\neq[/tex] F(x)) or (x [tex]\notin[/tex] domF and x [tex]\in[/tex] domG and y [tex]\neq[/tex] G(x))

I was hoping I could show that domF = domG and F(x) = G(x) for all x [tex]\in[/tex] domF was equivalent to this but I couldn't. This is as close as I could get:

(domF = domG and F(x) = G(x) for all x [tex]\in[/tex] domF) if and only if (x [tex]\in[/tex] domF if and only if x [tex]\in[/tex] domG and, if y [tex]\in[/tex] domF, then y = F(x) if and only if y = G(x))

(domF = domG and F(x) = G(x) for all x [tex]\in[/tex] domF) if and only if (((x [tex]\in[/tex] domF and x [tex]\in[/tex] domG and y = F(x) and y = G(x)) or (x [tex]\in[/tex] domF and x [tex]\in[/tex] domG and y [tex]\neq[/tex] F(x) and y [tex]\neq[/tex] G(x)) or (x [tex]\notin[/tex] domF and x [tex]\notin[/tex] domG))

This is close to what I had above but not quite.

I don't know. I think I need to get a set theory book that has solutions; I am currently working through "Introduction to Set Theory" by Hrbacek and Jech. I might switch to "Naive Set Theory" by Paul Hamos; it doesn't even have exercises I don't think.

Any help would be appreciated.
 
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  • #2
May I ask you how you defined "function"? Is this an exercise of Hrbacek and Jech, if so, which one? I need to see all the definitions before I can answer, since a function is probably something different for me...

On a related note, Hrbacek and Jech is a very good book, I really recommend it. Halmos is more about naive set theory, and is less formal. If you want to learn modern set theory, then Hrbacek and Jech is the best. If you just want to use the set theory somewhere, then I guess Halmos is easier...

Also I would like to add this reference: staff.science.uva.nl/~vervoort/AST/ast.pdf The theory in there is not at all easy, but it has some good and fun exercises (with solutions)
 
  • #3
Dont mind the above post, I checked Hrbacek and Jech and found it... For the record, I don't really like that definition...

So you indeed have

[tex] F=G~\text{iff}~ (xFy ~\Leftrightarrow~xGy)[/tex]

The domain of F is

[tex]dom(F)=\{x~\vert~\exists y:~xFy\}=\{x~\vert~\exists y:~xGy\}=dom(G) [/tex]

So the domains are equal. Now take an x in dom(F). Then there exists a y such that xFy=xGy. This means exactly that F(x)=G(x) (since y is unique).
 

1. What is the definition of a function in set theory?

In set theory, a function is a mathematical concept that describes the relationship between two sets, where each element in the first set (known as the domain) is mapped to a unique element in the second set (known as the codomain).

2. How is the equality of functions defined in set theory?

In set theory, two functions are considered equal if they have the same domain, the same codomain, and the same mapping of elements from the domain to the codomain.

3. Can two functions with different names be considered equal in set theory?

Yes, in set theory, the names of functions do not affect their equality. As long as the functions have the same domain, codomain, and mapping, they are considered equal.

4. How is the equality of functions different from the equality of sets in set theory?

The equality of sets in set theory is based on the elements within the sets being identical, while the equality of functions is based on the mapping of elements between sets being identical. In other words, two sets can be equal even if they have different elements, but two functions can only be equal if they have the same mapping.

5. Can functions with different domains or codomains be considered equal in set theory?

No, in set theory, functions with different domains or codomains are considered different functions, even if their mappings are identical. For example, a function with a domain of real numbers and a function with a domain of integers may have the same mapping, but they are still considered different functions.

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